Question

Write the following cubes in expanded form: ${\left(x-\frac{2y}{3}\right)}^3$

Answer 1

Expand the expression:

The given expression is, ${\left(x-\frac{2y}{3}\right)}^3$

We know the identity ${\left(a-b\right)}^3=a^3-3ab\left(a-b\right)-b^3$

Here, $a=x$ and $b=\frac{2y}{3}$

So,
$\begin{array}{rcl}{\left(x-\frac{2y}{3}\right)}^3& =& x^3-3\left(x\right)\left(\frac{2y}{3}\right)\left(x-\frac{2y}{3}\right)-{\left(\frac{2y}{3}\right)}^3\\ & =& x^3-2xy\left(x-\frac{2y}{3}\right)-\frac{8y^3}{27}\\ & =& x^3-2x^{2}y+\frac{4x{y}^2}{3}-\frac{8}{27}{y}^3\end{array}$

Thus, the expanded form of ${\left(x-\frac{2y}{3}\right)}^3$ is $\begin{array}{l}{x}^3-2x^{2}y+\frac{4x{y}^2}{3}-\frac{8}{27}{y}^3\end{array}$